The n-Category Café

put an inner product on a BC 2-vector space

I have changed my mind about the nicest way to put an inner product on the 2-space. It might be good to use categories internal to hermitean vector spaces instead of just plain vector spaces. Then a good notion of categorified inner product should be given by the internal $\mathrm{hom}$, making contact with the way I envisioned generalized Hilbert spaces of states here.

I have asked the experts and found the following answer to a crucial question in this BC $n$-vector business:

Question: We may regard an $n$-term chain complex as living in an $n$-category, whose morphisms are chain maps, 2-morphisms are chain homotopies, and so on.

Are equivalence classes in this $n$-category in bijection with cohomology classes, i.e. with chain complexes all whose maps are the 0-map?

Answer: yes, this is true. It is in fact true for chain complexes in any category of projective modules.

This means that whenever one sees people inverting “quasi isomorphisms” in the 1-category of chain complexes, this is a 1-categorical workaround for talking about equivalence in a setup that is really $\omega$-categorical.

Hence forming the derived category of chain complexes is a lot like working with the full $\omega$-category structure on chain complexes in the first place (i.e. with reagrding arbitrary chain complexes as $\omega$-vector spaces).

I gather that this is very well known to the experts. But I didn’t know it before.

Urs has a nice clear formulation here of the action of morphisms on 2-morphisms in the general linear 2-group of transformations on a BC 2-vector space. Perhaps we might consider possible sub-2-groups of this 2-group. Presumably we can form those which are full on 2-morphisms but which restrict independently on the two components of the 1-morphisms, e.g., to orthogonal groups.

On the other hand, if we go for non-full on 2-morphisms options, then we’ll have to be careful that the action of the 1-morphisms preserves them.

Turning my eyes regretfully down from the gods on the heights of Olympus, I shall continue ploughing my lonely furrow at its base.

I have been thinking about quotients recently. I think I understand some things better now, so I shall put down my observations in case they are helpful to anybody else.

Weak, firm and strict quotients

We start with a groupoid. We want to think of this as a ‘weak quotient’. This usually means the result of acting on a set with a group. For each member of the group that sends an element $a$ of the set to an element $b$, we add one morphism $a\overset{f}{\rightarrow}b.$ However, if, instead of considering an action (functor) from a group to Set, we consider more generally a functor from a groupoid to Set, sending all objects in the groupoid to the same set in Set, then we can think of any groupoid we like as being a weak quotient.

On the other hand, we have also the “strict” quotient, which is the set that we get by setting all isomorphic objects in the groupoid equal to one another and throwing away the morphisms (except identities).

I’d like to insert an intermediate level quotient. In this, we have neither a set of morphisms between two objects (as with the weak quotient), nor a “tautology” of morphisms (i.e. only the necessary identity morphisms) as with the strict quotient, but a truth value of morphisms. That is to say, if the weak quotient has at least one morphism between two objects, the intermediate quotient has exactly one. This is nothing other than a set together with an equivalence relation on it. This is nice, since an equivalence relation is traditionally seen as the intermediate step between a group action and a partition considered as a set.

What we are doing here is taking the obvious functor from the weak quotient to the strict quotient and factorising it into a full bit and a faithful bit.

Instead of this, we can try doing the steps in a different order. Instead of first reducing a set of morphisms to just one (per pair of objects), and then identifying objects, we can instead try identifying the objects first. To do this, we need to pick a particular isomorphism between each pair of objects, and “contract it down to the identity”, dragging the two objects together into one, and forcing the other isomorphisms and automorphisms to merge in a particular way. This gives us a set of groups. We can, if we wish, then force these down into a mere set of elements by identifying all automorphisms of a given object.

The fact that we can do this, merging at the 0-morphism level while failing to do so at the 1-morphism level, makes me wonder if perhaps we should have had a preliminary stage in which we add one 2-morphism from each 1-morphism to each other 1-morphism between the same objects, and then identified 1-morphisms by contraction of the 2-morphisms.

Quotients of groups

A group $G$ can act not only on an arbitrary set, but specifically on itself, by left multiplication. This gives the group a groupoid structure. The morphisms inherit a group operation from the group itself, and this turns the group into a 2-group $G\rightarrow G$, with the delta map being the identity, and the action being by conjugation.

We can also get a subgroup $H$ of $G$ to act on the whole group by multiplication. The resulting weak quotient is also a 2-group $H\rightarrow G$ in the same way, that is with the delta map the inclusion and the action upwards by conjugation. More interestingly, the intermediate quotient is the equivalence relation which partitions the elements into the various cosets of the subgroup, while the strict quotient is precisely the quotient of the group by its subgroup.

We can also homomorphically map some other group $K$ into a subgroup $H$ of the group $G$, and so long as the conjugation action of the whole group on the subgroup lifts correctly to the action of $G$ on $K,$ we get a general strict 2-group like this. So a strict 2-group is a generalised weak quotient of groups.

Parallel morphisms and boundaries

Time for a parenthetical comment. When we are identifying $k$-morphisms with each other for some $k,$ there is a condition that they should have the same source and target $k-1$-morphisms: that they should be “parallel”. How does this work for objects (0-morphisms)? Presumably they should have the same “-1-morphisms” as each other. What does this mean? Well, we can see by inspection that the actual condition is that the objects should belong to the same component of the groupoid. That’s the only way there can be 1-morphisms between them that can be contracted to the identity.

This is interesting, because in a cell complex, if the 0-cells are the vertices, then the -1-cells are the components, and the boundary of a vertex is the component it belongs to.

More generally, the condition that two $k$-morphisms should have the same source and target $k-1$-morphisms if they are to be identified, or, more generally, if they are to have any $k+1$-morphisms between them, amounts to saying that they should have the same boundary, where their boundary is some sort of formal difference of their source and target. In terms of the $k+1$-morphisms that putatively connect them, this amounts to saying that the boundaries of their source and target should be the same, i.e. have a difference of zero, or … OK, you’re way ahead of me … that the boundary of a boundary should be zero.

2-actions and 2-quotients

If a 1-action amounts to throwing in some morphisms to a set to form a groupoid, then a 2-action should amount to throwing in some 1-morphisms and 2-morphisms between them. Obviously, the 2-group (or 2-groupoid) that is the source of the action should send its own 1-morphisms to the new 1-morphisms and its 2-morphisms to the new 2-morphisms.

If the acted-upon object is a not merely a set, but a groupoid, we expect the 2-action to be by functors and natural isomorphisms between them. The result is, I think, an internal groupoid in the (or a) 2-category of groupoids.

More specifically, a 2-group $H\overset{\delta}{\rightarrow}G$ can act on itself. The action of $G$ is on itself is by multiplication, and its action on $H$ is the one given in the definition of the 2-group. In addition, the elements of $Ker(\delta)$ can act on $H$ by multiplication. The restriction to $Ker(\delta)$ is to ensure that the resulting 2-morphisms only send morphisms in $H$ to each other if they go between the same objects (i.e. if they differ by an automorphism of those objects, i.e. a member of $Ker(\delta)$.)

We can then repeat the same process as before, taking a subgroup of $Ker(\delta)$ or homomorphically mapping another group $K$ into it, to get a 3-group $K\overset{\delta_1}{\rightarrow}H\overset {\delta_0}{\rightarrow}G$.

The condition that the image of $\delta_1$ lies in the kernel of $\delta_0$, or, in effect, that $\delta^2=0$ is of course precisely required by the condition that the boundary of a boundary be zero.

I have more things to say, but I’m not feeling too well just at the moment, so that’s all I can manage for now.

As always, apologies if I have been

a) too obvious, or
b) too obscure, or
c) both

On another thread I mention an introduction to stratified spaces by Paul Gunnells. He describes there what is the Schubert stratification of a Grassmanian, in particular $G(4, 2)$. So you take a maximal flag, i.e., a sequence of subspaces of each dimension, each containing its predecessor. Then you see how $R^2$ can intersect with this flag. It will do so in a non-decreasing sequence of natural numbers $(0, a, b, c, 2)$, where each entry is at most 1 more than its predecessor. Here $(a, b, c)$ can be (0, 0, 1), (0, 1, 1), (0, 1, 2), (1, 1, 1), (1, 1, 2), or (1, 2, 2). For each of these there is a Schubert cell of members of the Grassmanian, which fit together to make $G(4, 2)$.

Now something we’d like to know is the structure of the Grassmanian of $R^{a,b}$ sub vector 2-spaces of $R^{p,q}$. So presumably we need to choose a maximal flag, which in this case is not just a sequence of inclusions, but a lattice, with one entry for each pair of natural numbers no greater than $(p, q)$. Then we need to see how $(a, b)$ subspaces can intersect with this flag.

Even small dimensional examples produce large numbers of Schubert cells. Take the Grassmanian $G((2, 2), (1, 1))$. A maximal flag has the shape of a diamond lattice with 9 entries. Intersection freedom is limited to 5 places, and in 4 of these only freedom in one of the dimensions. With a very quick calculation, there look to be 18 Schubert cells.